System of two Equations: $\sqrt x+y=11$ and $\sqrt y+x=7$

Question

A friend of Mine gave me a system of two equations and asked me to solve them $\rightarrow$

$$\sqrt{x}+y=11~~ ...1$$ $$\sqrt{y}+x=7~~ ...2$$

I tried to solve them manually and got this horrendously complicated fourth degree equation $\rightarrow$

$$\begin{align*} y &= (7-x)^2 ~...\mbox{(from 2)} \\ y &= 49 - 14 x + x^2 \\ \implies 11&= \sqrt{x}+ 49 - 14 x + x^2 ...(\mbox{from 1)}\\ \implies~~ 0&=x^4-28x^3+272x^2-1065x+1444 \end{align*}$$

Solving this wasn't exactly my piece of cake but I could tell that one of Solutions would have been 9 and 4

But my friend kept asking for a formal solution.

I tried plotting the equations and here's what I got $\rightarrow$

enter image description here

So the equations had two pairs of solutions (real ones).

Maybe, Just maybe I think these could be solved using approximations.

So How do i solve them using a formal method (Calculus,Algebra,Real Analysis...)

P.S. I'm In high-school.

From the fourth degree equation, you can use Ferrari's method (http://en.wikipedia.org/wiki/Quartic_function) to solve it. — guaraqe, Jun 01 '12 at 15:05
In this question a solution for the same system (with $x,y$ interchanged) is asked. As such the present question is a duplicate. $$\sqrt{x} + y = 7$$ $$\sqrt{y} + x =11.$$ — Américo Tavares, Jun 01 '12 at 15:41
@AméricoTavares I didnt know about it being asked earlier.....meaning my question is going to be closed.......damn i worked hard asking this and prettying it up!! — The-Ever-Kid, Jun 01 '12 at 15:44
@AméricoTavares: Is there a difference in that this OP wants a formal solution? And the other question actually doesn't give that.\ — Thomas, Jun 01 '12 at 15:46
@Thomas: My answer to the second question I've linked to in my second comment above gives an algebraic solution. — Américo Tavares, Jun 01 '12 at 17:03
@thomas some of your confusion on Robert's answer below may be stemming from the graph you used. I'm not sure what you used to graph the equations, but it has plotted both branches of the square roots. If you were to plot the initial system of equations implicitly, a la: http://i.imgur.com/ayQ6m.jpg you can see there's only one real solution — Andrew Christianson, Jun 01 '12 at 17:24
@DrewChristianson i was the one who asked the question and graphed the function. I used mathematica BTW you used gnuplot didnt you? — The-Ever-Kid, Jun 01 '12 at 18:29
@the-ever-kid sorry, I got confused with the names. I'm not sure what you're getting at with your comment about gnuplot. Just because yours is from mathematica doesn't make it more right (mine was from Maple btw). Just to prove it to you though, here's the same implicit graph from the original system of equations from Mathematica: http://imgur.com/dPqq1 which you can draw yourself with: $$ContourPlot[{y + (x)^{(1/2)} == 11 , x + (y)^{(1/2)} == 7}, {x, 0, 10}, {y, 0, 10}]$$
also, wolframalpha: http://www.wolframalpha.com/input/?i=plot+y%2Bsqrt%28x%29%3D11+%2C+x%2Bsqrt%28y%29%3D7 — Andrew Christianson, Jun 01 '12 at 18:56
@DrewChristianson Calm down dude.. I'm not sure what you used to graph the equations that's what you said so i told you what i used. Just because yours is from mathematica doesn't make it more right I know pal, mine is incorrect cause it was my error...i solved for y in the second equation hence the error. You seriously need to calm down.. I'm not sure what you're getting at with your comment about gnuplot. I guessed gnuplot, 'cause gnuplot always plots the first eq with red the other with green..anyways thanks — The-Ever-Kid, Jun 02 '12 at 04:41
@the-ever-kid Sorry, I was a little snarky. I thought you were contending your graph was correct because it was from mathematica and you thought mine was from gnuplot. I think we've both been reassured that tone does not always come through well on the internet. — Andrew Christianson, Jun 02 '12 at 17:27
@DrewChristianson I think we've both been reassured that tone does not always come through well on the internet. Not really 'cause in real life we never get second chances to clarify. If one gets pissed off that's it end of story . At least here you get to clarify your intent/tone. — The-Ever-Kid, Jun 03 '12 at 04:54
This is the same system as in http://math.stackexchange.com/questions/108962/ — Martin Sleziak, Dec 14 '13 at 07:43

score 6 · Accepted Answer · answered Jun 01 '12 at 15:29

6

Assume $x$ and $y$ are integers. Notice that, in this case, if $\sqrt x +y=11$, an integer, then $\sqrt x $ must be an integer. A similar argument can be made for $y$. So if they're integers then they're both perfect squares. Rephrasing in terms of the square roots (still integers) $X=\sqrt x,Y=\sqrt y$ $$X+Y^2=11$$ $$Y+X^2=7$$ subtracting the second equation from the first: $$X-Y+Y^2-X^2=4$$ $$(X-Y)+(Y-X)(Y+X)=4$$ $$(Y-X)(X+Y-1)=4$$ Both of the brackets are integers, so the only values they can take are the factors of $4$. So either $$Y-X=2,X+Y-1=2$$ or $$Y-X=4,X+Y-1=1$$ or$$Y-X=1,X+Y-1=4$$ Solving each of these is simple. The only one that gives positive integer values (the conditions of our little set up here) is the $3^{rd}$ one, which gives the answer you found. Keep in mind that there's nothing wrong with guessing and playing around with the problem first, then coming to a more structured argument later. If you want a full analytic solution you could use the quartic equation on the one you have and rule out the other solutions as involving the wrong branches of $\sqrt x$, but it's pointlessly messy.

answered Jun 01 '12 at 15:29

Robert Mastragostino

15,727

How come the graphs intersect twice and the points other than 4,9 are not integers..... – The-Ever-Kid Jun 01 '12 at 15:35
Mathematica plots figured two solutions other than 4,9 -> 9.8051180869527 and 7.8686874817494 – The-Ever-Kid Jun 01 '12 at 15:41
it's the bit about branches of $\sqrt x$ i mentioned at the end. $\sqrt x$ is defined as the positive number whose square is $x$. So $\sqrt 4=x$, gives $x=2$. But $x^2=4$ gives us two solutions, $\pm 2$. The extra solution isn't part of our original problem and not something we care about. Rephrasing your question in terms of squares and not square roots introduces extra 'solutions' that don't solve your original problem. The extra answer mathematica produced only works if you take the negative square root for the second equation, which isn't how the square root function is defined. – Robert Mastragostino Jun 01 '12 at 16:16
look at the image that is a perfectly real and positive graph – The-Ever-Kid Jun 01 '12 at 16:35
the solutions arnt negative – The-Ever-Kid Jun 01 '12 at 16:35
right. $x$ and $y$ (which you found) are not negative. But plug them into the equations. They work for the first one, not for the second one. What does work for the second one and your numbers is $x-\sqrt y=7$. I.e. taking $-\sqrt y$. So you're correct in saying $y$ isn't negative. But it's the same thing as taking the negative square root of $4$ as $-2$. $4$ itself being positive doesn't change that. – Robert Mastragostino Jun 01 '12 at 16:58

score 2 · Answer 2 · answered Jun 01 '12 at 17:29

Once you guessed the solutions, you can easily prove that there are no others. Rewrite the equations as $y=11-\sqrt x=F(x)$ and $x=7-\sqrt y=G(y)$. Note that both $x,y\le 11$, so their square roots are at most $4$, which means that $x,y\ge 3$. Now just observe that $z\mapsto \sqrt z$ is a contraction on $[3,\infty)$ (the difference of values is less than the difference of arguments). Thus, $F$ and $G$ are also contractions whence if we had two different solutions $(x_1,y_1)$ and $(x_2,y_2)$, we would get $$ |x_1-x_2|=|G(y_1)-G(y_2)|<|y_1-y_2|=|F(x_1)-F(x_2)|<|x_1-x_2| $$ which is absurd.

System of two Equations: $\sqrt x+y=11$ and $\sqrt y+x=7$

2 Answers2